Bindings in helium

• Binding energy
• Example
• Test

Binding energy

Theory of atomic nucleuses has proved that binding energy between particular nucleons does not depend on their charge, i.e. that it approximately holds:
E(N,N)=E(N,P)=E(P,P), where N is neutron and P proton.

Often reported reason is, that nucleouses with inverse number of protons and neutrons (e.g. izotopes H(3,1),(p=1,n=2) and He(3,2),(p=2,n=1)) are similar and their bindings energy differs only by Coulomb repulsion of protons.

Now let us do one test.

Example

Let us assume, we can compute bindings energy with this formula:

E(p,n) = c(P,P)*E(P,P) + c(P,N)*E(P,N) + c(N,N)*E(N,N)

where p,n is number of protons P and neutrons N and c(P,P), c(P,N), c(N,N) number of bindings between them.

Bindings energy of the four simplest atomic nucleuses is:

  X(A,Z)   p  n   E(p,n)  c(P,P) c(P,N) c(N,N)
---------------------------------------------
  H (2,1)  1  1   2.224    0      1      0
  H (3,1)  1  2   8.482    1      2      0
  He(3,2)  2  1   7.718    0      2      1
  He(4,2)  2  2  28.295    1      4      1

We get three equations:

  E(1,1) = 0*E(P,P) + 1*E(P,N) + 0*E(N,N) = 2.224
  E(1,2) = 0*E(P,P) + 2*E(P,N) + 1*E(N,N) = 8.482
  E(2,1) = 1*E(P,P) + 2*E(P,N) + 0*E(N,N) = 7.718

Hence:

	 E(P,P) = 3.270
	 E(N,N) = 4.034
	 E(P,N) = 2.224

Check:
	 E(1,1) =  1*2.224           = 2.224
	 E(1,2) =  2*2.224 + 1*4.034 = 8.482
	 E(2,1) =  2*2.224 + 1*3.270 = 7.718

Binding energy of E(2,2) is then:

	 E(2,2) = 1*E(P,P) + 4*E(P,N) + 1*E(N,N)
	 E(2,2) = 1*3.270  + 4*2.224  + 1*4.034 = 16.200.

It is - in comparison with real energy of He(4,2) - very low
(the difference 28.295- 16.200= 12.095 is c. 3*E(N,N)).

Test

Let us complete the formula for E(p,n):

E(p,n) = c(P,P)*E(P,P)+c(P,N)*E(P,N)+c(N,N)*E(N,N)+p*E(P)+n*E(N),

where

  E(1,1) = 0*E(P,P) + 1*E(P,N) + 0*E(N,N) +1*E(P)+ 1*E(N) = 2.224
  E(1,2) = 0*E(P,P) + 2*E(P,N) + 1*E(N,N) +1*E(P)+ 2*E(N) = 8.482
  E(2,1) = 1*E(P,P) + 2*E(P,N) + 0*E(N,N) +2*E(P)+ 1*E(N) = 7.718
  E(2,2) = 1*E(P,P) + 4*E(P,N) + 1*E(N,N) +2*E(P)+ 2*E(N) =28.295

If E(N)=E(P) we get:

	  E(P)   = -6.048
	  E(N)   = -6.048
	  E(P,P) = -2.778
	  E(N,N) = -2.014
	  E(P,N) = 14.319

Check:
	 E(1,1) = 14.319 -6.048 - 6.048 = 2.224
	 E(1,2) = 2*14.319 -2.014 -3* 6.048 = 8.482
	 E(2,1) = -2.778 + 2* 14.319 -3*6.048 = 7.718
	 E(2,2) = -2.778 + 4* 14.319 -2.014 -4*6.048  = 28.295

The binding energy E(P,P) is nearly the same as E(N,N), but not the same as energy E(P,N).
Nucleouses with inverse number of protons and neutrons confirm or disprove nothing - they have the same number of bindings E(P,N).